[amirkdv]

> you can show [...] every linear differential operator [that commutes with translations and rotations] is a polynomial in the Laplacian.

I may be tripping over something here but this doesn't sound right. If you mean polynomial in the real number sense, i.e.

Lu = a0 + a1 Delta(u) + a2 Delta(u)^2 + ...

(where Delta = Laplacian, and a0, a1, ... real numbers), then is this true? The famous wave operator doesn't have this form.

And if you mean "polynomial" as in a series over function space, i.e.

Lu ~ a0 + a1 Du + a2 D^2u + ... (infinite terms, not equality but convergence)

where D is the usual differential operator and a0 is a number, a1 is a 1-d vector, a2 is a 2d matrix, so on, then this is standard calculus of variations on any reasonable function space. Taylor series for function spaces if you want. I don't think that's limited to nice Galilean operators.

[woopwoop]

Thanks for pointing this out, I should have been more precise. I was referring to time invariant equations. You are right that this doesn't apply to the wave operator (which also does not commute with rotations in R^4). But in space only, yes polynomial means polynomial with constant coefficients.

The proof basically goes that commuting with translations implies immediately that L has constant coefficients. Then on the Fourier side applying L translates to multiplying by a polynomial, and commuting with rotations translates to the claim that that polynomial is rotation invariant. And every rotation invariant polynomial in several variables is p(|x|^2) for some polynomial p in on variable. Then on the spacial side p(|x|^2) translates to L = p(laplacian).